Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to use the Rayleigh-Ritz method to calculate an approximate solution to the extremum problem with the following functional: $$ L[y]=\int\int_D (u_x^2+u_y^2+u^2-2xyu)\,dx\,dy, $$ $D$ is the unit square i.e. $0 \leq x \leq 1, 0 \leq y \leq 1.$ Also $u=0$ on the boundary of $D$.

I have chosen to use the trial function: $$ \phi(x,y)=cxy(1-x)(1-y) $$ Where $c$ is a constant that I need to find.

I am familiar with using the Rayleigh-Ritz method most of the time, however this question I am not sure of. Is it possible to convert the problem to a Sturm-Liouville ration type?

Thanks for your help.

share|cite|improve this question

Your integral is in the form of $$L(x,y,u)=\int\int_D (u_x^2+u_y^2+u^2-2xyu)\,dx\,dy$$ $$0 \leq x \leq 1, 0 \leq y \leq 1$$ Due to homogenous boundary conditions it is possible to use your approximation function $$u(x,y)=cxy(1-x)(1-y)$$ When substituted into integral equation $$L(x,y,u)=\int_0^1\int_0^1 (u_x^2+u_y^2+u^2-2xyu)\,dx\,dy=\frac{7}{300}c^2-\frac{1}{72}c$$ and taking first derivative condition and solving for c $$\frac{d\,L}{d\,c}=\frac{7}{150}c-\frac{1}{72}=0\Rightarrow c=\frac{25}{84}$$ and $$u(x,y)=\frac{25}{84}xy(1-x)(1-y)$$ Since the second derivative is positive it is a minimum.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.